Question

Find all possible values of y for which the number 53y1 is divisible by 3. Also, find each such number.

Answer 1

If a number is divisible by 3, then the sum of the digits is also divisible by 3.

Sum of the digits = $5+3+y+1=9+y$

The sum of the digits is divisible by 3 in the following cases:

$$\begin{gathered} 9+y=9,ory=0 \\ \mathrm{Then}\mathrm{the}\mathrm{number}\mathrm{is}5301. \\ 9+y=12,ory=3 \\ \mathrm{Then}\mathrm{the}\mathrm{number}\mathrm{is}5331. \\ 9+y=15,ory=6 \\ \mathrm{Then}\mathrm{the}\mathrm{number}\mathrm{is}5361. \\ 9+y=18,ory=9 \\ \mathrm{Then}\mathrm{the}\mathrm{number}\mathrm{is}5391. \end{gathered}$$

∴ y = 0, 3, 6 or 9
The possible numbers are 5301, 5331, 5361 and 5391.